Implicit differentiation

Differentiation is possible without solving for $y$ first¹. We find the tangent to the circle $x^2 + y^2 = 25$ at the point $(3, 4)$ .

Differentiating both sides

Since $y$ is a function of $x$ , the chain rule makes the derivative of $y^2$ equal to $2y \dfrac{dy}{dx}$ .

\begin{align*} 2x + 2y \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= -\frac{x}{y} \end{align*}

Putting in $(3, 4)$ gives the slope $-\dfrac{3}{4}$ . The tangent is therefore as follows.

y = -\frac{3}{4}(x - 3) + 4 = -\frac{3}{4}x + \frac{25}{4}

Compared with solving first

Method	Expression	What it covers
Solving for $y$	$y = \pm\sqrt{25 - x^2}$	upper and lower semicircles separately
Leaving it implicit	$\dfrac{dy}{dx} = -\dfrac{x}{y}$	both at once, in one formula

Differentiating implicitly handles the two halves together.

The geometric meaning

The radius drawn from the center $(0, 0)$ to $(3, 4)$ has slope $\dfrac{4}{3}$ , and the tangent has slope $-\dfrac{3}{4}$ ; the product is $-1$ . The fact that a tangent is perpendicular to the radius falls out of the formula.

Where it cannot be used

It fails where $y = 0$ . At $(5, 0)$ and $(-5, 0)$ the denominator vanishes, but that is because the tangent there is the vertical line $x = 5$ or $x = -5$ , which has no slope. The way the formula breaks matches the geometry.

The same for curves other than circles

Curve	Differentiated form	Example slope
$x^2 + y^2 = 25$	$-\dfrac{x}{y}$	$-\dfrac{3}{4}$ at $(3, 4)$
$x^3 + y^3 = 9$	$-\dfrac{x^2}{y^2}$	$-\dfrac{1}{4}$ at $(1, 2)$
$xy = 6$	$-\dfrac{y}{x}$	$-\dfrac{3}{2}$ at $(2, 3)$

Even for a curve that cannot be solved for $y$ , the tangent can still be found. A product mixed in changes nothing: for $xy = 6$ the product rule gives $y + x\dfrac{dy}{dx} = 0$ , agreeing with the $-\dfrac{6}{x^2}$ obtained by differentiating $y = \dfrac{6}{x}$ .

The two arcs on the graph are the circle $x^2 + y^2 = 25$ , the line is the tangent, and the large dots are the point of tangency $(3, 4)$ and the center $(0, 0)$ .

Implicit function, Wikipedia